5
$\begingroup$

Among separable metrizable spaces:

Cantor set is the unique compact zero-dimensional space without isolated points.

$\mathbb Q$ is the unique countable space without isolated points

$\mathbb R \setminus \mathbb Q$ is the unique zero-dimensional, $G_\delta$-space with no compact neighborhood.

$\mathbb Q ^\omega$ is the unique zero- dimensional, first category $F_{\sigma\delta}$-space with the property that no nonempty clopen subset is a $G_{\delta\sigma}$-space.

Question. Is there a simple classification of zero-dimensional $G_{\delta\sigma}$-spaces which have no compact neighborhoods? The simplest examples would be $\mathbb Q$, $\mathbb R\setminus \mathbb Q$, and $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$. Are there many others?

Is there a nice characterization of $\mathbb Q\times (\mathbb R\setminus \mathbb Q)$?

$\endgroup$
4
$\begingroup$

This paper by Van Mill from 1981 gives a characterisation of $\Bbb Q \times \Bbb P$ (where $\Bbb P$ is a common notation for the irrationals) in Thm 5.3:

If $X$ is separable metrisable and zero-dimensional, $\sigma$-complete and nowhere complete and nowhere $\sigma$-compact then $X \simeq \Bbb Q \times \Bbb P$.

Where by complete I mean topologically complete (i.e. in this context: completely metrisable) and a nowhere-$P$ space is one where no non-empty open subset has property $P$, so $\Bbb P$ and $\Bbb Q$ are nowhere locally compact, e.g.)

I think $\Bbb Q \times C$ (with $C$ the Cantor set) is another example for your list.

A lot of information can be found in van Engelen's PhD-thesis from 1985: homogeneous zero-dimensional absolute Borel sets, where he shows there are are $\omega_1$ many homeomorphism types of subsets of $C$ that are both $F_{\sigma \delta}$ and $G_{\delta\sigma}$, also separately written up here. In his thesis he also gave the first characterisation of $\Bbb Q^\omega$ (now relegated to the appendix of it). Look up Van Engelen's work from around that time for related results.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.